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WA4. Differentiation Exercises

Question 1

1. Chains Inc. is in the business of making and selling chains.

  • Let c(t) be the number of miles of chain produced after t hours of production.
  • Let p© be the profit as a function of the number of miles of chain produced
  • Let q(t) be the profit as a function of the number of hours of production.
  • Suppose the company can produce 3 miles of chain per hour and suppose their profit on the chains is $4000 per mile of chain.

1. Find and interpret (use complete sentences) each of the following (include units), c’(t) , p’© , and q’(t)?

  • c’(t) = 3 miles per hour, that is, the company produces at a constant rate = 3 miles of chain per hour.
  • p’© = 4000 dollars per mile, that is, the company makes 4000 dollars of profit per mile of chain produced.
  • q’(t) = 12000 dollars per hour, that is, the company produces 3 miles per hour and makes 4000 of profit for each mile of chain produced, thus it makes 12000 dollars per hour.

2. How does q’(t) relates to p’© and c’(t) ?.

q’(t) = p’© * c’(t) = 4000 * 3 = 12000

The q(t) represents the profit per hour, which in turn is the product of the profit per mile and the number of miles produced per hour.

Question 2

2. Use Desmos to graph the function \(y^3+yx^2+x^2-3y^2=0\) and estimate the slope of the tangent line at (-1,1). Then find \(\frac{dy}{dx}\) using implicit differentiation and plug in x=-1 and y=1 . Compare and discuss the estimated slope with the slope you found analytically.

The graph of the function is shown below:

Well, I’m not sure how to estimate the slope of the tangent line at (-1,1) using Desmos, but it definitely must be negative as it looks from the graph above.

Now, let’s find \(\frac{dy}{dx}\) using implicit differentiation:

\[ \begin{aligned} y^3+yx^2+x^2-3y^2 &= 0 \\ \frac{d}{dx}(y^3+yx^2+x^2-3y^2) &= \frac{d}{dx}(0) \\ \frac{d}{dx}(y^3) + \frac{d}{dx}(yx^2) + \frac{d}{dx}(x^2) - \frac{d}{dx}(3y^2) &= 0 \\ 3y^2 \frac{dy}{dx} + (x^2 \frac{dy}{dx} + 2yx) + 2x - 6y \frac{dy}{dx} &= 0 \\ \frac{dy}{dx} (3y^2 + x^2 - 6y) &= -2x - 2yx \\ \frac{dy}{dx} &= \frac{-2x - 2yx}{3y^2 + x^2 - 6y} \\ \end{aligned} \]

The slope at (-1,1) equals \(\frac{dy}{dx}\) when x=-1 and y=1:

\[ \begin{aligned} \frac{dy}{dx} &= \frac{-2(-1) - 2(-1)(1)}{3(1)^2 + (-1)^2 - 6(1)} \\ &= \frac{4}{-2} \\ &= -2 \end{aligned} \]

And the slope of the tangent line at (-1,1) is -2, which is negative as expected.

The tangent equation is: \(y = -2x - 1\) and it is plotted with the graph of the function below:

Question 3

3. Let \(f(x)=(3x^2+1)^2\). Find \(f'(x)\) in 3 different ways by following the instructions below in parts a, b and c:

a) Develop the identity f(x) then take the derivative.

\[ \begin{aligned} f(x) &= (3x^2+1)^2 \\ &= (3x^2+1)(3x^2+1) \\ &= 9x^4 + 3x^2 + 3x^2 + 1 \\ &= 9x^4 + 6x^2 + 1 \\ f'(x) &= 36x^3 + 12x \\ &= 12x(3x^2 + 1) \end{aligned} \]

b) View f(x) as \((3x^2+1)(3x^2+1)\) and use the product rule to find \(f'(x)\).

The product rule states that if \(f(x) = g(x)h(x)\), then \(f'(x) = g'(x)h(x) + g(x)h'(x)\), so:

\[ \begin{aligned} f(x) &= (3x^2+1)(3x^2+1) \\ f'(x) &= (3x^2+1)'(3x^2+1) + (3x^2+1)(3x^2+1)' \\ &= (6x)(3x^2+1) + (3x^2+1)(6x) \\ &= (3x^2+1)(6x + 6x) \\ &= (3x^2+1)(12x) \\ &= 12x(3x^2+1) \end{aligned} \]

c) Apply the chain rule directly to the expression \(f(x)=(3x^2+1)^2\) .

The chain rule states that if \(f(x) = g(h(x))\), then \(f'(x) = g'(h(x))h'(x)\), so we can consider \(f(x) = g(h(x))\) where \(g(x) = x^2\) and \(h(x) = 3x^2 + 1\):

\[ \begin{aligned} g(x) &= x^2 \\ g'(x) &= 2x \\ h(x) &= 3x^2 + 1 \\ h'(x) &= 6x \\ \end{aligned} \]

And let’s apply the chain rule:

\[ \begin{aligned} f(x) &= g(h(x)) \\ f'(x) &= g'(h(x))h'(x) \\ &= 2(3x^2 + 1)6x \\ &= 12x(3x^2 + 1) \end{aligned} \]

d) Are your answers in parts a, b, c the same? Why or why not?.

All answers are the same as we are differentiating the same function just using different methods, so the result must be the same.

Question 4

4. Find \(\frac{dy}{dx}\) for the equation \(3y^2-cos(y)=x^3\).

\[ \begin{aligned} 3y^2-cos(y) &= x^3 \\ 3y^2 - x^3 - cos(y) = 0 \\ \frac{d}{dx}(3y^2 - x^3 - cos(y)) &= \frac{d}{dx}(0) \\ \frac{d}{dx}(3y^2) - \frac{d}{dx}(x^3) - \frac{d}{dx}(cos(y)) &= 0 \\ 6y \frac{dy}{dx} - 3x^2 + sin(y) \frac{dy}{dx} &= 0 \\ \frac{dy}{dx} (6y + sin(y)) &= 3x^2 \\ \frac{dy}{dx} &= \frac{3x^2}{6y + sin(y)} \end{aligned} \]

Question 5

5. Find the equation of the tangent line that passes through point (1,2) to the graph \(8y^3+x^2y-x=65\).

Let’s compute \(\frac{dy}{dx}\) first:

\[ \begin{aligned} 8y^3+x^2y-x &= 65 \\ \frac{d}{dx}(8y^3+x^2y-x) &= \frac{d}{dx}(65) \\ \frac{d}{dx}(8y^3) + \frac{d}{dx}(x^2y) - \frac{d}{dx}(x) &= 0 \\ 24y^2 \frac{dy}{dx} + (x^2 \frac{dy}{dx} + 2xy) - 1 &= 0 \\ \frac{dy}{dx} (24y^2 + x^2) &= 1 - 2xy \\ \frac{dy}{dx} &= \frac{1 - 2xy}{24y^2 + x^2} \end{aligned} \]

Now, let’s find the slope of the tangent line at (1,2) by plugging in x=1 and y=2 into \(\frac{dy}{dx}\):

\[ \begin{aligned} \frac{dy}{dx} &= \frac{1 - 2xy}{24y^2 + x^2} \\ &= \frac{1 - 2(1)(2)}{24(2)^2 + (1)^2} \\ &= \frac{1 - 4}{96 + 1} \\ &= \frac{-3}{97} \end{aligned} \]

Now, let’s find the equation of the tangent line using the slope and the point (1,2):

\[ \begin{aligned} y - y_1 &= m(x - x_1) \\ y - 2 &= \frac{-3}{97}(x - 1) + c \\ \end{aligned} \]

we need another point to find c, let’s use, but I used c=2 and graphed the function and the tangent line below:

Question 6

6. Find \(\frac{dy}{dx}\) for the equation \(xy^2-x^2y=2\).

\[ \begin{aligned} xy^2-x^2y &= 2 \\ \frac{d}{dx}(xy^2-x^2y) &= \frac{d}{dx}(2) \\ \frac{d}{dx}(xy^2) - \frac{d}{dx}(x^2y) &= 0 \\ y^2 + 2xy \frac{dy}{dx} - 2xy - x^2 \frac{dy}{dx} &= 0 \\ \frac{dy}{dx} (2xy - x^2) &= 2xy - y^2 \\ \frac{dy}{dx} &= \frac{2xy - y^2}{2xy - x^2} \end{aligned} \]

Question 7

7. Find \(f'(x)\) for the function \(f(x)= \sqrt[4]{-3x^4-2}\).

We will use the chain rule to find \(f'(x)\), considering \(f(x) = g(h(x))\) where

\[ \begin{aligned} g(x) &= \sqrt[4]{x} \\ h(x) &= -3x^4 - 2 \\ g'(x) &= \frac{1}{4}x^{-\frac{3}{4}} \\ h'(x) &= -12x^3 \\ \end{aligned} \]

and now, let’s find \(f'(x)\):

\[ \begin{aligned} f(x) &= g(h(x)) \\ f'(x) &= g'(h(x)) . h'(x) \\ &= \frac{1}{4}(-3x^4 - 2)^{-\frac{3}{4}} . (-12x^3) \\ &= \frac{-12x^3}{4} \frac{1}{(-3x^4 -2)^\frac{3}{4}} \\ &= -3x^3 \frac{1}{(-3x^4 -2)^\frac{3}{4}} \\ &= \frac{-3x^3}{\sqrt[4]{(-3x^4 -2)^3}} \\ \end{aligned} \]

Question 8

8. Find \(f'(x)\) for the function \(f(x)=(5x^2+3)^4\).

We will use the chain rule to find \(f'(x)\), considering \(f(x) = g(h(x))\) where

\[ \begin{aligned} g(x) &= x^4 \\ h(x) &= 5x^2 + 3 \\ g'(x) &= 4x^3 \\ h'(x) &= 10x \\ \end{aligned} \]

and now, let’s find \(f'(x)\):

\[ \begin{aligned} f(x) &= g(h(x)) \\ f'(x) &= g'(h(x)) . h'(x) \\ &= 4(5x^2 + 3)^3 . 10x \\ &= 40x(5x^2 + 3)^3 \\ \end{aligned} \]

Question 9

9. Find \(f'(x)\) for the function \(f(x)=sin^2(cos(4x))\).

We will use the chain rule to find \(f'(x)\), considering \(f(x) = g(h(x))\) where

\[ \begin{aligned} g(x) &= sin^2(x) \\ h(x) &= cos(4x) \\ g'(x) &= 2sin(x)cos(x) \\ h'(x) &= -4sin(4x) \\ \end{aligned} \]

and now, let’s find \(f'(x)\):

\[ \begin{aligned} f(x) &= g(h(x)) \\ f'(x) &= g'(h(x)) . h'(x) \\ &= 2sin(cos(4x))cos(cos(4x)) . -4sin(4x) \\ &= -8sin(cos(4x))sin(4x)cos(cos(4x)) \\ \end{aligned} \]

Question 10

10. Find \(f'(x)\) for the following functions:

a) \(f(x) = cos(ln4x^3))\)

We will use the chain rule to find \(f'(x)\), considering \(f(x) = g(h(x))\) where

\[ \begin{aligned} g(x) &= cos(x) \\ h(x) &= ln(4x^3) \\ g'(x) &= -sin(x) \\ h'(x) &= (ln (4x^3))' = (ln' (4x^3)).(4x^3)' \\ &= \frac{1}{4x^3} . 12x^2 \\ &= \frac{3}{x} \\ \end{aligned} \]

and now, let’s find \(f'(x)\):

\[ \begin{aligned} f(x) &= g(h(x)) \\ f'(x) &= g'(h(x)) . h'(x) \\ &= -sin(ln(4x^3)) . \frac{3}{x} \\ &= -\frac{3sin(ln(4x^3))}{x} \\ \end{aligned} \]

b) \(f(x)= e^{(4x^3+5)^2}\)

We will use the chain rule to find \(f'(x)\), considering \(f(x) = g(h(x))\) where

\[ \begin{aligned} g(x) &= e^x \\ h(x) &= (4x^3+5)^2 \\ g'(x) &= e^x \\ h'(x) &= ((4x^3+5)(4x^3+5))' \\ &= (4x^3+5)'(4x^3+5) + (4x^3+5)(4x^3+5)' \\ &= 12x^2(4x^3+5) + (4x^3+5)12x^2 \\ &= 24x^2(4x^3+5) \\ \end{aligned} \]

and now, let’s find \(f'(x)\):

\[ \begin{aligned} f(x) &= g(h(x)) \\ f'(x) &= g'(h(x)) . h'(x) \\ &= e^{(4x^3+5)^2} . 24x^2(4x^3+5) \\ &= 24x^2(4x^3+5)e^{(4x^3+5)^2} \\ \end{aligned} \]